Ebook Higher engineering mathematics (5th edition) Part 2

(BQ) Part 2 book Higher engineering mathematics has contents: Differentiation of parametric equations, differentiation of implicit functions, partial differentiation, standard integration, integration using partial fractions, some applications of integration,...and other contents.

Certain mathematical functions can be expressed

more simply by expressing, say, x and y separately

in terms of a third variable For example, y = r sin θ,

x = r cos θ Then, any value given to θ will produce

a pair of values for x and y, which may be plotted to

provide a curve of y = f (x).

The third variable, θ, is called a parameter and the

two expressions for y and x are called parametric

equations.

The above example of y = r sin θ and x = r cos θ

are the parametric equations for a circle The

equa-tion of any point on a circle, centre at the origin and

of radius r is given by: x2+ y2= r2, as shown in

Chapter 14

To show that y = r sin θ and x = r cos θ are suitable

parametric equations for such a circle:

Left hand side of equation

= r2= right hand side

(since cos2θ+ sin2θ= 1, as shown in

Chapter 16)

29.2 Some common parametric

equations

The following are some of the most common

param-etric equations, and Figure 29.1 shows typical shapes

(f) Astroid x = a cos3θ, y = a sin3θ

(g) Cycloid x = a (θ− sin θ) , y = a (1− cos θ)

dx dθ

= 2(2θ− 1)

2

5(2θ− 1)

Problem 2 The parametric equations of a

function are given by y = 3 cos 2t, x = 2 sin t.

Determine expressions for (a)dy



dx dt

Problem 3 The equation of a tangent drawn to

a curve at point (x1, y1)is given by:

y − y1= dy1

dx1(x − x1)Determine the equation of the tangent drawn to

the parabola x = 2t2, y = 4t at the point t.

Problem 4 The parametric equations of a

cycloid are x = 4(θ − sin θ), y = 4(1 − cos θ).

Determine (a) dy

dx (b)

d2y

dx2

dx dθ

Now try the following exercise

Exercise 128 Further problems on

differen-tiation of parametric equations

1 Given x = 3t − 1 and y = t(t − 1), determine

3 The parametric equations for an ellipse

are x = 4 cos θ, y = sin θ Determine (a) dy

The equation of a tangent drawn to a curve at

point (x1, y1) is given by:

y − y1=dy1

dx1(x − x1)

Use this in Problems 6 and 7

6 Determine the equation of the tangent drawn

to the ellipse x = 3 cos θ, y = 2 sin θ at θ = π

6.

[y = −1.155x + 4]

7 Determine the equation of the tangent drawn

to the rectangular hyperbola x = 5t, y =5

Problem 5 The equation of the normal drawn

to a curve at point (x1, y1)is given by:

y − y1= −dy1

1

dx1(x − x1)

Determine the equation of the normal drawn to

the astroid x= 2 cos3θ, y= 2 sin3θat the point

Problem 6 The parametric equations for a

hyperbola are x = 2 sec θ, y = 4 tan θ Evaluate



dx dθ



= −cos3θ

sin3θ = − cot3θ When θ= 1 rad, d2y

dx2= − cot31= − 1

(tan 1)3

= −0.2647, correct to 4 significant figures.

Problem 7 When determining the surface

ten-sion of a liquid, the radius of curvature, ρ, of part

of the surface is given by:

ρ=

567

1+



dy dx

2 3

d2y

dx2

Find the radius of curvature of the part of the

surface having the parametric equations x = 3t2,



dx dt

Now try the following exercise

Exercise 129 Further problems on

differen-tiation of parametric equations

1 A cycloid has parametric equations

x = 2(θ − sin θ), y = 2(1 − cos θ).

Eval-uate, at θ = 0.62 rad, correct to 4 significant

The equation of the normal drawn to

a curve at point (x1, y1) is given by:

y − y1= −dy1

1

dx1(x − x1)

Use this in Problems 2 and 3

2 Determine the equation of the normal drawn

3 Find the equation of the normal drawn to the

cycloid x = 2(θ − sin θ), y = 2(1 − cos θ) at

6rad for the cardioid

x = 5(2θ − cos 2θ), y = 5(2 sin θ − sin 2θ).

[0.02975]

5 The radius of curvature, ρ, of part of a

sur-face when determining the sursur-face tension of

a liquid is given by:

When an equation can be written in the form y = f (x)

it is said to be an explicit function of x Examples

of explicit functions include

y = 2x3− 3x + 4, y = 2x ln x

and y= 3ex

cos x

In these examples y may be differentiated with

respect to x by using standard derivatives, the

prod-uct rule and the quotient rule of differentiation

respectively

Sometimes with equations involving, say, y and x,

it is impossible to make y the subject of the formula.

The equation is then called an implicit function

and examples of such functions include

y3+ 2x2= y2− x and sin y = x2+ 2xy.

30.2 Differentiating implicit functions

It is possible to differentiate an implicit function

by using the function of a function rule, which may

stitution u = y3 is made, from which, du

dy = 3y2.Hence, d

dx (y

3)= (3y2)×dy

dx, by the function of a

function rule

A simple rule for differentiating an implicit

func-tion is summarised as:

d

dx [ f ( y)]= d

dy [ f ( y)]×dy

Problem 1 Differentiate the following

func-tions with respect to x:

(b) Let u = sin 3t, then, by the function of a function

Problem 2 Differentiate the following

func-tions with respect to x:

320 DIFFERENTIAL CALCULUS

Now try the following exercise

Exercise 130 Further problems on

differen-tiating implicit functions

In Problems 1 and 2 differentiate the given

func-tions with respect to x.

(c) −2

et

dt dy

The product and quotient rules of differentiation

must be applied when differentiating functions

con-taining products and quotients of two variables

Now try the following exercise.

Exercise 131 Further problems on

differen-tiating implicit functions involving products

30.4 Further implicit differentiation

An implicit function such as 3x2+ y2− 5x + y = 2,

may be differentiated term by term with respect to

using equation (1) and standard derivatives

An expression for the derivative dy

Thus when x = 4 and y = ±3, dy

dx= − 4

±3= ±

4 3

322 DIFFERENTIAL CALCULUS

x2+ y2= 25 is the equation of a circle, centre at

the origin and radius 5, as shown in Fig 30.1 At

x= 4, the two gradients are shown

y

5 3

Gradient

= 4 3

x2 + y2 = 25

Figure 30.1

Above, x2+ y2= 25 was differentiated implicitly;

actually, the equation could be transposed to

y=(25− x2) and differentiated using the function

of a function rule This gives

Problem 9 Find the gradients of the tangents

drawn to the circle x2+ y2− 2x − 2y = 3 at

x= 2

The gradient of the tangent is given by dy

dx Differentiating each term in turn with respect to x

4

x2 +y2 −2x−2y= 3

= 1 2

= − 1 2

Figure 30.2

The circle having the given equation has its centre at

(1, 1) and radius√

5 (see Chapter 14) and is shown

in Fig 30.2 with the two gradients of the tangents

Problem 10 Pressure p and volume v of a gas

are related by the law pv γ = k, where γ and k

are constants Show that the rate of change of

pressure dp

dt = −γ p

v

dv dt

i.e dp

dt = −γ p

v

dv dt

Now try the following exercise

Exercise 132 Further problems on implicit differentiation

In Problems 1 and 2 determine dy

9 Determine the gradients of the tangents

drawn to the circle x2+ y2= 16 at the point

where x= 2 Give the answer correct to 4

11 Determine the gradient of the curve

3xy + y2= −2 at the point (1,−2) [−6]

With certain functions containing more complicated

products and quotients, differentiation is often made

easier if the logarithm of the function is taken before

differentiating This technique, called ‘logarithmic

differentiation’ is achieved with a knowledge of

(i) the laws of logarithms, (ii) the differential

coef-ficients of logarithmic functions, and (iii) the

differ-entiation of implicit functions

31.2 Laws of logarithms

Three laws of logarithms may be expressed as:

(i) log(A × B) = log A + log B

(ii) log



A B



= log A − log B

(iii) log A n = n log A

In calculus, Napierian logarithms (i.e logarithms to

a base of ‘e’) are invariably used Thus for two

func-tions f (x) and g(x) the laws of logarithms may be

The differential coefficient of the logarithmic

func-tion ln x is given by:

(x+ 2) may be achieved by using the

product and quotient rules of differentiation; ever the working would be rather complicated Withlogarithmic differentiation the following procedure

(iii) Differentiate each term in turn with respect to x

using equations (1) and (2)

by law (iii) of Section 31.2

(iii) Differentiating with respect to x gives:

Using logarithmic differentiation and following theprocedure gives:

Now try the following exercise

Exercise 133 Further problems on tiating logarithmic functions

differen-In Problems 1 to 6, use logarithmic ation to differentiate the given functions withrespect to the variable

Whenever an expression to be differentiated

con-tains a term raised to a power which is itself a

function of the variable, then logarithmic

differen-tiation must be used For example, the

differentia-tion of expressions such as x x , (x+ 2)x,√x

ln y = ln x x = x ln x, by law (iii) of Section 31.2

Differentiating both sides with respect to x gives:

by law (iii) of Section 31.2

Differentiating each side with respect to x gives:

i.e ln y = (3x + 2) ln x, by law (iii) of Section 31.2

Differentiating each term with respect to x gives:

Now try the following exercise

Exercise 134 Further problems on

differen-tiating [ f (x)] xtype functions

In Problems 1 to 4, differentiate with respect to x

The marks for each question are shown in

brackets at the end of each question.

1 Differentiate the following with respect to the

variable:

(a) y= 5 + 2√x3− 1

x2 (b) s= 4e2θ sin 3θ (c) y=3 ln 5t

cos 2t (d) x= 2

(t2− 3t + 5) (13)

2 If f (x) = 2.5x2− 6x + 2 find the co-ordinates at

the point at which the gradient is−1 (5)

3 The displacement s cm of the end of a stiff spring

at time t seconds is given by:

s = ae −kt sin 2πft Determine the velocity and

acceleration of the end of the spring after

2 seconds if a = 3, k = 0.75 and f = 20. (10)

4 Find the co-ordinates of the turning points on the

curve y = 3x3+ 6x2+ 3x − 1 and distinguish

5 The heat capacity C of a gas varies with absolute

temperature θ as shown:

C = 26.50 + 7.20 × 10−3θ − 1.20 × 10−6θ2

Determine the maximum value of C and the

temperature at which it occurs (5)

6 Determine for the curve y = 2x2− 3x at the point

(2, 2): (a) the equation of the tangent (b) the

7 A rectangular block of metal with a square section has a total surface area of 250 cm2 Findthe maximum volume of the block of metal (7)

cross-8 A cycloid has parametric equations given by:

x = 5(θ − sin θ) and y = 5(1 − cos θ) Evaluate

(a) dy

dx (b)

d2y

dx2 when θ = 1.5 radians Give

answers correct to 3 decimal places (8)

9 Determine the equation of (a) the tangent, and

(b) the normal, drawn to an ellipse x = 4 cos θ,

y = sin θ at θ = π

10 Determine expressions for dz

dy for each of the

Differential calculus

32

Differentiation of hyperbolic functions

32.1 Standard differential coefficients of

Using the quotient rule of differentiation the

deriva-tives of tanh x, sech x, cosech x and coth x may be

determined using the above results

Problem 1 Determine the differential

coeffi-cient of: (a) th x (b) sech x.

cosech ax −a cosech ax coth ax

coth ax −a cosech2ax

Problem 4 Differentiate the following with

respect to the variable: (a) y = 4 sin 3t ch 4t

Now try the following exercise

Exercise 135 Further problems on tiation of hyperbolic functions

differen-In Problems 1 to 5 differentiate the given tions with respect to the variable:

Differential calculus

33

Differentiation of inverse trigonometric and hyperbolic functions

Inverse trigonometric functions are denoted by

prefixing the function with ‘arc’ or, more

com-monly, by using the −1 notation For example, if

y = sin x, then x = arcsin y or x = sin−1y Similarly,

if y = cos x, then x = arccos y or x = cos−1y, and so

on In this chapter the−1 notation will be used A

sketch of each of the inverse trigonometric functions

is shown in Fig 33.1

Inverse hyperbolic functions are denoted by

pre-fixing the function with ‘ar’ or, more commonly, by

using the−1notation For example, if y = sinh x, then

x = arsinh y or x = sinh−1y Similarly, if y = sech x,

then x = arsech y or x = sech−1y, and so on In this

chapter the −1 notation will be used A sketch of

each of the inverse hyperbolic functions is shown in

Fig 33.2

33.2 Differentiation of inverse

trigonometric functions

(i) If y= sin−1x, then x = sin y.

Differentiating both sides with respect to y

(ii) A sketch of part of the curve of y= sin−1x

is shown in Fig 33(a) The principal value ofsin−1x is defined as the value lying between

−π/2 and π/2 The gradient of the curve

between points A and B is positive for all values

of x and thus only the positive value is taken

when evaluating √ 1

1− x2.(iii) Given y= sin−1 x

(iv) Given y= sin−1 f (x) the function of a function

rule may be used to find dy

dx.

Figure 33.1

0 1 2 3 x

1

3 2

y

x

y = cosh  1 x

1 2 3

(v) The differential coefficients of the remaining

inverse trigonometric functions are obtained in

a similar manner to that shown above and a

summary of the results is shown in Table 33.1

Table 33.1 Differential coefficients of inverse

(a) If y= cos−1x then x = cos y.

Differentiating with respect to y gives:

thus the differential coefficient dy

a and show that the

differ-ential coefficient of tan−12x

The principal value of y= tan−1x is defined as

the angle lying between −π

4+ x2

=

32

Let u= cos−13x then y = ln u.

By the function of a function rule,

336 DIFFERENTIAL CALCULUS

Problem 7 Differentiate y = x cosec−1x.

Using the product rule:

Now try the following exercise

Exercise 136 Further problems on differentiating inverse trigonometric functions

In Problems 1 to 6, differentiate with respect tothe variable

1 (a) sin−14x (b) sin−1 x

Inverse hyperbolic functions may be evaluated most

conveniently when expressed in a logarithmic

From Chapter 5, ey = cosh y + sinh y and

cosh2y− sinh2y= 1, from which,

cosh y=1+ sinh2y which is positive since cosh y

is always positive (see Fig 5.2, page 43)

Hence ey=1+ sinh2y + sinh y

= 1+x

a

2+ x

a = a2+ x2

a2

+ x

= 1.4436, correct to 4 decimal places

Problem 10 Show thattanh−1 x

and evaluate, correct

to 4 decimal places, tanh−13

= 0.6931, correct to 4 decimal places

Problem 11 Prove that

dy = a cosh y (from Chapter 32).

Also cosh2y− sinh2y= 1, from which,

From the sketch of y= sinh−1 x shown in

Fig 33.2(a) it is seen that the gradient

The remaining inverse hyperbolic functions are

differentiated in a similar manner to that shown

above and the results are summarized in Table 33.2

Problem 12 Find the differential coefficient

34

2

− x2

=

349

16− x2

=

34

=−(1 − x2)

(1− x2) + 2x tanh−1x = 2x tanh−1x − 1

3 tanh

−1 2x

3 + c

Now try the following exercise

Exercise 138 Further problems on tiation of inverse hyperbolic functions

differen-In Problems 1 to 11, differentiate with respect tothe variable

34.1 Introduction to partial derivatives

In engineering, it sometimes happens that the

varia-tion of one quantity depends on changes taking place

in two, or more, other quantities For example, the

volume V of a cylinder is given by V = πr2h The

volume will change if either radius r or height h

is changed The formula for volume may be stated

mathematically as V = f (r, h) which means ‘V is

some function of r and h’ Some other practical

examples include:

(i) time of oscillation, t = 2π l

g i.e t = f (l, g).

(ii) torque T = Iα, i.e T = f (I, α).

(iii) pressure of an ideal gas p=mRT

V i.e p = f (T, V).

(iv) resonant frequency f r= 1

2π√

LC i.e f r = f (L, C), and so on.

When differentiating a function having two

vari-ables, one variable is kept constant and the

dif-ferential coefficient of the other variable is found

with respect to that variable The differential

coef-ficient obtained is called a partial derivative of

the function

34.2 First order partial derivatives

A ‘curly dee’, ∂, is used to denote a differential

coef-ficient in an expression containing more than one

∂h are examples of first order partial

derivatives, since n= 1 when written in the form

∂ n V

∂r n.First order partial derivatives are used when findingthe total differential, rates of change and errors forfunctions of two or more variables (see Chapter 35),when finding maxima, minima and saddle points forfunctions of two variables (see Chapter 36), and withpartial differential equations (see Chapter 53)

Problem 5 Pressure p of a mass of gas is given

by pV = mRT, where m and R are constants,

V is the volume and T the temperature Find

pen-g where l is the length

of the pendulum and g the free fall acceleration

due to gravity Determine∂t

Now try the following exercise.

Exercise 139 Further problems on first

order partial derivatives



t + c sin



nπb L

34.3 Second order partial derivatives

As with ordinary differentiation, where a

differen-tial coefficient may be differentiated again, a pardifferen-tial

derivative may be differentiated partially again to

give higher order partial derivatives

(i) Differentiating∂V

∂r of Section 34.2 with respect

to r, keeping h constant, gives ∂

second order partial derivatives.

(vi) It is seen from (iii) and (iv) that ∂

no sudden jumps or breaks)

Second order partial derivatives are used in thesolution of partial differential equations, in waveg-uide theory, in such areas of thermodynamics cov-ering entropy and the continuity theorem, and whenfinding maxima, minima and saddle points for func-tions of two variables (see Chapter 36)

Problem 7 Given z = 4x2y3− 2x3+ 7y2find(a) ∂

Now try the following exercise.

Exercise 140 Further problems on second order partial derivatives

In Problems 1 to 4, find (a) ∂

In Chapter 34, partial differentiation is introduced for

the case where only one variable changes at a time,

the other variables being kept constant In practice,

variables may all be changing at the same time

If z = f (u, v, w, ), then the total differential,

dz, is given by the sum of the separate partial

y + 1, determine the total differential, dz.

The total differential is the sum of the partial

Problem 2 If z = f (u, v, w) and z = 3u2−

2v + 4w3v2find the total differential, dz.

The total differential





pV T

 dp +  p

pV T

 dV

i.e dT = T

p dp+ T

V dV

Now try the following exercise

Exercise 141 Further problems on the total

6 If z = f (a, b, c) and z = 2ab − 3b2c + abc,

find the total differential, dz.



b(2 + c) da + (2a − 6bc + ac) db + b(a − 3b) dc



7 Given u = ln sin (xy) show that

du = cot (xy)(y dx + x dy)

35.2 Rates of change

Sometimes it is necessary to solve problems in which

different quantities have different rates of change

From equation (1), the rate of change of z, dz

correct to 4 significant figures

Problem 5 The height of a right circular cone

is increasing at 3 mm/s and its radius is ing at 2 mm/s Determine, correct to 3 significantfigures, the rate at which the volume is chang-ing (in cm3/s) when the height is 3.2 cm and theradius is 1.5 cm

decreas-Volume of a right circular cone, V=1

2ac sin B, where B is the angle between

sides a and c If a is increasing at 0.4 units/s, c

is decreasing at 0.8 units/s and B is increasing at

0.2 units/s, find the rate of change of the area of

the triangle, correct to 3 significant figures, when

a is 3 units, c is 4 units and B is π/6 radians.

Using equation (2), the rate of change of area,

Problem 7 Determine the rate of increase of

diagonal AC of the rectangular solid, shown in

Fig 35.1, correct to 2 significant figures, if the

sides x, y and z increase at 6 mm/s, 5 mm/s and

4 mm/s when these three sides are 5 cm, 4 cmand 3 cm respectively

Now try the following exercise.

Exercise 142 Further problems on rates of

change

1 The radius of a right cylinder is

increas-ing at a rate of 8 mm/s and the height is

decreasing at a rate of 15 mm/s Find the

rate at which the volume is changing in

cm3/s when the radius is 40 mm and theheight is 150 mm [+226.2 cm3/s]

2 If z = f (x, y) and z = 3x2y5, find the rate of

change of z when x is 3 units and y is 2 units when x is decreasing at 5 units/s and y is

increasing at 2.5 units/s [2520 units/s]

3 Find the rate of change of k, correct to

4 significant figures, given the following

data: k = f (a, b, c); k = 2b ln a + c2e a ; a is increasing at 2 cm/s; b is decreasing at

3 cm/s; c is decreasing at 1 cm/s; a = 1.5 cm,

b = 6 cm and c = 8 cm. [515.5 cm/s]

4 A rectangular box has sides of length x cm,

y cm and z cm Sides x and z are expanding

at rates of 3 mm/s and 5 mm/s respectively

and side y is contracting at a rate of 2 mm/s.

Determine the rate of change of volume when

x is 3 cm, y is 1.5 cm and z is 6 cm.

[1.35 cm3/s]

5 Find the rate of change of the total surfacearea of a right circular cone at the instantwhen the base radius is 5 cm and the height

is 12 cm if the radius is increasing at 5 mm/sand the height is decreasing at 15 mm/s

[17.4 cm2/s]

35.3 Small changes

It is often useful to find an approximate value forthe change (or error) of a quantity caused by smallchanges (or errors) in the variables associated with

the quantity If z = f (u, v, w, ) and δu, δv, δw,

denote small changes in u, v, w, respectively,

then the corresponding approximate change δz in

z is obtained from equation (1) by replacing the

differentials by the small changes

mine the approximate percentage error in k when

the pressure is increased by 4% and the volume

is decreased by 1.5%

i.e the approximate error in k is a 1.9% increase.

Problem 9 Modulus of rigidity G = (R4θ)/L,

where R is the radius, θ the angle of twist and L

the length Determine the approximate

percent-age error in G when R is increased by 2%, θ is

reduced by 5% and L is increased by 4%.



(0.02R)+



R4L

100G

Hence the approximate percentage error in G is

a 1% decrease.

Problem 10 The second moment of area of

a rectangle is given by I = (bl3)/3 If b and l

are measured as 40 mm and 90 mm respectivelyand the measurement errors are−5 mm in b and +8 mm in l, find the approximate error in the

an error of 0.2% too large and g 0.1% too small.

Using equation (3), the approximate change in t,

Problem 12 Find the differential coefficient

34

2< /small>... x2< /small>

=

349

16− x2< /small>

=

34

=−(1 − x2< /sup>)

(1− x2< /small>) + 2x...

cm3/s when the radius is 40 mm and theheight is 150 mm [ +22 6 .2 cm3/s]

2 If z = f (x, y) and z = 3x2< /small>y5, find the rate of

change